Maths Olympiad Prep

Track / Stage 4 / 298 of 340 #558 of 1964

Problem 558

AMC 12 late, AIME early
Number theory Difficulty 5.0 Prove it

Let aa be an integer. Show that 5 divides a2a^{2} if and only if 5 divides aa.

We can slightly generalize Euclid's lemma:

Theorem 5 (Generalized Euclid's Lemma).

Let pp be a prime and a1,a2,,ana_{1}, a_{2}, \ldots, a_{n} be integers. If pp divides a1a2ana_{1} \cdot a_{2} \cdots a_{n}, then it divides one of a1,a2,,ana_{1}, a_{2}, \ldots, a_{n}.

Proof. We will prove this generalization by applying Euclid's lemma multiple times. If a prime pp divides a1a2ana_{1} \cdot a_{2} \cdots a_{n}, then by Euclid's lemma, pp divides either a1a_{1} or a2a3ana_{2} \cdot a_{3} \cdot \ldots a_{n}. In the first case, we are done. In the second case, we apply Euclid's lemma again in the same manner. We then have that pp divides either a2a_{2} or a3a4ana_{3} \cdot a_{4} \cdots a_{n}. In the first case, we are done; in the second case, we repeat the same reasoning. If we repeat this reasoning enough times, we will eventually be done.

## Remark 6.

This reasoning can be made more rigorous using induction.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

5 is prime. Therefore, by Euclid's lemma, if 5 divides a2=aaa^{2}=a \cdot a, then either 5 divides aa, or 5 divides aa, so we have concluded.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.